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2 years ago

A. How far does a 100-newton force have to move to do 1,000 joules

Physics

1 answer:

Aloiza [94]2 years ago

5 0

Work done by a force is given as the product of force and the distance moved by the force.

Work done is the product of force and the distance moved by the the force.

Work done = Force × distance

Thus, distance required by the 100 N force is given as:

Distance = work done/force

Distance = 1000/100 = 10 m

Distance to be moved is 10 m.

Force applied = work done/ distance

Force applied by the hoist = 500/2

Force applied by the hoist = 250 N

Distance moved in one push-up = 25 cm = 0.25 m

Work done by the athlete after one push-up = 250 × 0.25 m

Work done by the athlete = 62.5 J

Distance moved by the force = 0 m

Work done = 500 × 0 = 0 N

Therefore, for work to be done, force has to move a distance.

Learn more about work done at: brainly.com/question/25573309

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A crate is pushed up each of the two ramps shown in the diagram below. Based on the concept that simple machines make work easie

Naddik [55]

Answer:

ramp b requires less force than ramp a

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7 0

3 years ago

Can anyone solve these for my by using unit vectors? Can you also please show your work

Oxana [17]

4. The Coyote has an initial position vector of $\vec r_0=(15.5\,\mathrm m)\,\vec\jmath$ .

4a. The Coyote has an initial velocity vector of $\vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath$ . His position at time $t$ is given by the vector

$\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2$

where $\vec a$ is the Coyote's acceleration vector at time $t$ . He experiences acceleration only in the downward direction because of gravity, and in particular $\vec a=-g\,\vec\jmath$ where $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$ . Splitting up the position vector into components, we have $\vec r=r_x\,\vec\imath+r_y\,\vec\jmath$ with

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$r_y=15.5\,\mathrm m-\dfrac g2t^2$

The Coyote hits the ground when $r_y=0$ :

$15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s$

4b. Here we evaluate $r_x$ at the time found in (4a).

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m$

5. The shell has initial position vector $\vec r_0=(1.52\,\mathrm m)\,\vec\jmath$ , and we're told that after some time the bullet (now separated from the shell) has a position of $\vec r=(3500\,\mathrm m)\,\vec\imath$ .